Problem: Let the operation $m^{**}n$ be defined as $m^{**}n=(m-n)^2 \div (m+n)^2$, for all real values of $m$ and $n$. What is the value of $100^{**}50$? Express your answer as a common fraction.
Solution: We plug in $100$ for $m$ and $50$ for $n$. Therefore, $100^{**}50=(100-50)^2 \div (100+50)^2 = \dfrac{50^2}{150^2} = \dfrac{50^2}{3^2 \cdot 50^2} = \dfrac{1}{3^2} = \boxed{\dfrac19}$.